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ivann1987 [24]
3 years ago
13

X+3y-2z=10, -x-2y+z=-7, 3x+9y-5z=28

Mathematics
2 answers:
NISA [10]3 years ago
6 0
Hmm
what you do is try to eliminate 1 variable in 2 equaions first
we can elminate x's in first and 2nd equation

add first and 2nd equation
x+3y-2z=10
<u>-x-2y+z=-7 +</u>
0x+y-z=3
y-z=3


multiply 2nd equation by 3 and add to last one
-3x-6y+3z=-21
<u>3x+9y-5z=28 +</u>
0x+3y-2z=7

3y-2z=7

we now have
y-z=3
3y-2z=7
multiply first equation by -2 and add to 2nd

-2y+2z=-6
<u>3y-2z=7 +</u>
y+0z=1
y=1

now we can sub back
y-z=3
1-z=3
minus 1
-z=2
times -1
z=-2

sub baack into any equation
x+3(1)-2(-2)=10
x+3+4=10
x+7=10
minus 7
x=3

x=3
y=1
z=-2
(3,1,-2)

Semenov [28]3 years ago
4 0
  
\displaystyle  \\ &#10;~~~~x+3y-2z=10 ~~~~~(E_1)\\  &#10;-x-2y+z=-7 ~~~~~(E_2)\\ &#10;~~3x+9y-5z=28~~~~~(E_3) \\ &#10;\text{- - - - - } \\ &#10;E_1 + E_2 ~~\Longrightarrow~~y - z = 3 \\  \\ &#10;-3 \times E_1  + E3  ~~\Longrightarrow~~ z = -2 \\  \\ &#10;\Longrightarrow\Longrightarrow\Longrightarrow \\  \\ &#10;y-z=3\\&#10;z = \boxed{-2}\\&#10;\text{- - - - - }\\&#10;y-(-2)=3\\&#10;y+2=3\\&#10;y =3-2=\boxed{1}\\&#10;\text{- - - - - }\\&#10;x+3y-2z=10\\&#10;x+3\cdot 1-2\cdot (-2)=10\\&#10;x+3+4=10\\&#10;x+7=10\\&#10;x= 10 - 7 = \boxed{3}



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